Find
∂
2
𝑓
∂
𝑦
∂
𝑥
∂y∂x
∂
2
f
​
for
𝑓
(
𝑥
,
𝑦
)
=
cos
⁡
(
𝑥
2
−
𝑦
3
)
f(x,y)=cos(x
2
−y
3
).

To find the mixed second partial derivative \(\frac{\partial^2 f}{\partial y \partial x}\) for \( f(x,y) = \cos(x^2 - y^3) \), we follow these steps:

1. **First Partial Derivative with respect to \( x \)**:
   \[
   f(x,y) = \cos(x^2 - y^3)
   \]
   Let \( u = x^2 - y^3 \). Then \( f(x,y) = \cos(u) \).

   \[
   \frac{\partial f}{\partial x} = \frac{d}{du} \cos(u) \cdot \frac{\partial u}{\partial x}
   \]
   \[
   \frac{d}{du} \cos(u) = -\sin(u) 
   \]
   \[
   \frac{\partial u}{\partial x} = \frac{\partial}{\partial x} (x^2 - y^3) = 2x
   \]
   Therefore,
   \[
   \frac{\partial f}{\partial x} = -\sin(x^2 - y^3) \cdot 2x = -2x \sin(x^2 - y^3)
   \]

2. **Second Partial Derivative with respect to \( y \)**:
   Now, we take the partial derivative of \(\frac{\partial f}{\partial x}\) with respect to \( y \).

   \[
   \frac{\partial}{\partial y} \left( -2x \sin(x^2 - y^3) \right)
   \]

   We apply the chain rule again:
   \[
   \frac{\partial}{\partial y} \left( -2x \sin(x^2 - y^3) \right) = -2x \cdot \frac{d}{du} \sin(u) \cdot \frac{\partial u}{\partial y}
   \]

   \[
   \frac{d}{du} \sin(u) = \cos(u)
   \]
   \[
   \frac{\partial u}{\partial y} = \frac{\partial}{\partial y} (x^2 - y^3) = -3y^2
   \]

   Therefore,
   \[
   \frac{\partial}{\partial y} \left( -2x \sin(x^2 - y^3) \right) = -2x \cos(x^2 - y^3) \cdot (-3y^2)
   \]

   \[
   \frac{\partial^2 f}{\partial y \partial x} = 6xy^2 \cos(x^2 - y^3)
   \]

So, the mixed second partial derivative is:
\[
\frac{\partial^2 f}{\partial y \partial x} = 6xy^2 \cos(x^2 - y^3)
\]

Would you like more details on any part of this solution or have any questions?

Here are 5 questions you can ask next:
1. How do you find higher-order partial derivatives for more complex functions?
2. Can you explain the concept of mixed partial derivatives?
3. How do you apply the chain rule in multivariable calculus?
4. What are some common mistakes to avoid when computing partial derivatives?
5. How do you interpret the physical meaning of partial derivatives in applied problems?

**Tip**: Always verify if the function is continuous and differentiable in the domain of interest before applying partial differentiation rules.

Learn how to find the mixed second partial derivative ∂²f/∂y∂x for the function f(x,y) = cos(x² - y³). This problem involves multivariable calculus, partial derivatives, and the application of the chain rule. Suitable for undergraduate students.

Math Problem Statement

Find ∂ 2 𝑓 ∂ 𝑦 ∂ 𝑥 ∂y∂x ∂ 2 f for 𝑓 ( 𝑥 , 𝑦 )

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